$\text{Đáp án+Giải thích các bước giải:}$

$\text{Bài 1:}$

$\text{2NaOH + $H_{2}SO_{4}$→$Na_{2}SO_{4}$ + $2H_{2}O$}$

$\text{Ta có: $n_{H_{2}SO_{4}}$=0,1.0,5= 0,05 (mol)}$

$\text{Theo PT: $n_{NaSO_{4}}$=$n_{H_{2}SO_{4}}$= 0,05 (mol)}$

$\text{⇒ $m_{NaSO_{4}}$=0,05.142= 7,1 (g)}$

$\text{Bài 2:}$

$\text{Gọi n là hóa trị của kim loại R}$

$\text{2R + $nH_{2}SO_{4}$→$R_{2}(SO_{4})_{n}$ + $nH_{2}$}$

$\text{Ta có: $n_{R}$=$\dfrac{2,52}{M_{R}}$}$

$\text{$n_{R_{2}(SO_{4})_{n}}$=$\dfrac{6,84}{M_{R}.2 + (32 + 16.4)n}$}$

$\text{Theo PT: $n_{R}$=2$n_{H_{2}SO_{4}}$}$

$\text{⇔ $\dfrac{2,52}{M_{R}}$ = 2 × $\dfrac{6,84}{M_{R}.2 + (32 + 16.4)n}$}$

$\text{⇔ $\dfrac{2,52}{M_{R}}$ = $\dfrac{13,68}{M_{R}.2 + (32 + 16.4)n}$}$

$\text{⇔ 2,52($2M_{R}$ + 96n )= 13,68$M_{R}$}$

$\text{⇔ 5,04$M_{R}$ + 241,92n = 13,68$M_{R}$}$

$\text{⇔ 241,92n = 13,68$M_{R}$ - 5,04$M_{R}$= 8,64$M_{R}$}$

$\text{⇒ $M_{R}$=$\dfrac{241,92n}{8,64}$= 28n}$

$\text{Xét: n= 1 → $M_{R}$ = 28 (loại)}$

$\text{n= 2 → $M_{R}$ = 56 (Fe)}$

$\text{n= 3 → $M_{R}$ = 84 (loại)}$

$\text{Vậy R là kim loại Fe có hóa trị II}$

Em tham khảo nha:

\(\begin{array}{l}
1)\\
2NaOH + {H_2}S{O_4} \to N{a_2}S{O_4} + 2{H_2}O\\
{n_{{H_2}S{O_4}}} = 0,1 \times 0,5 = 0,05\,mol\\
{n_{N{a_2}S{O_4}}} = {n_{{H_2}S{O_4}}} = 0,05\,mol\\
{m_{N{a_2}S{O_4}}} = 0,05 \times 142 = 7,1g\\
2)\\
2R + n{H_2}S{O_4} \to {R_2}{(S{O_4})_n} + n{H_2}\\
{n_R} = 2{n_{{R_2}{{(S{O_4})}_n}}} \Leftrightarrow \dfrac{{2,52}}{{{M_R}}} = 2 \times \dfrac{{6,84}}{{2{M_R} + 96n}}\\
 \Rightarrow {M_R} = 28n\,g/mol\\
n = 2 \Rightarrow {M_R} = 56g/mol \Rightarrow R:Fe
\end{array}\)